CODEWITHSUNDEEP
c
Jung Jae Won
Jung Jae Won

Reputation: 57

How can I search for the number in the array?

  1. I have to get user input and search for the user input in the array
  2. If there is the number that is exactly matching with user input then I have to display "The number you entered is in the array"
  3. If there is no number matching in array then I'm required to display "The number you entered is not in the array"

but currently I am experiencing the wrong output, I have double checked that user input was in the array but the result displayed as "no number matching".

How can I solve this problem?

    #include <stdio.h>
    #include <stdlib.h>
    #include <time.h>

    void arrayresult(int arraylength[], int &input_num)
    {   
        for (int i = 0; i < 10; i++)
        {
            if (arraylength[i] == input_num)
            {
                printf("The number you entered is in the array");
                break;
            }

            else
            {
                printf("The number you entered is not in the array");
                break;
            }
        }
    }

    int main(void)
    {
        int input_num;
        int arrayindex;
        int arraylength[10];
        int i = 0;

        srand(time(NULL));

        for (i = 0; i < 10; i++)
        {
            arrayindex += 1;
            arraylength[i] = rand() % (99 + 1 - 11) + 11;
            printf("A[%d]: %d ", arrayindex, arraylength[i]);
        }


        printf("\n");
        printf("Enter the number you want to search: ");
        scanf("%d", &input_num);

        arrayresult(arraylength, input_num);

        return 0;
    }

Upvotes: 1

Views: 89

Answers (5)

VJAYSLN
VJAYSLN

Reputation: 483

Simply,Your program tend to check only the first index of the arraylength with that input number. If its fails you are out of the for loop. Time to know about the uses of flag variable. If Flag is set, the number is already found, and don't wants to check again. If Flag is unset, the number is not found. 

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

void arrayresult(int arraylength[], int input_num)
{   
    int flag = 0;
    for (int i = 0; i < 10; i++)
    {
        if (arraylength[i] == input_num)
        {
            printf("The number you entered is in the array");
            flag = 1;
            break;
        }


    }
if (flag == 0)
        {
            printf("The number you entered is not in the array");
        }
}

int main(void)
{
    int input_num;
    int arrayindex = 0;
    int arraylength[10];
    int i = 0;

    srand(time(NULL));

    for (i = 0; i < 10; i++)
    {
        arrayindex += 1;
        arraylength[i] = rand() % (99 + 1 - 11) + 11;
        printf("A[%d]: %d ", arrayindex, arraylength[i]);
    }


    printf("\n");
    printf("Enter the number you want to search: ");
    scanf("%d", &input_num);

    arrayresult(arraylength, input_num);

    return 0;
}

Hope this might helps : )

Upvotes: 0

BrianChen
BrianChen

Reputation: 183

This code might help you.

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

void arrayresult(int arraylength[], int input_num)
{   
    for (int i = 0; i < 10; i++)
    {
        if (arraylength[i] == input_num)
        {
            printf("The number you entered is in the array");
            return;
        }
    }
    printf("The number you entered is not in the array");
}

int main(void)
{
    int input_num;
    int arraylength[10];
    int i = 0;

    srand(time(NULL));

    for (i = 0; i < 10; i++)
    {
        arraylength[i] = rand() % (99 + 1 - 11) + 11;
        printf("A[%d]: %d ", i + 1, arraylength[i]);
    }

    printf("\n");
    printf("Enter the number you want to search: ");
    scanf("%d", &input_num);

    arrayresult(arraylength, input_num);

    return 0;
}

Upvotes: 0

ratsafalig
ratsafalig

Reputation: 502

    void arrayresult(int arraylength[], int &input_num)
    {   
        for (int i = 0; i < 10; i++)
        {
            if (arraylength[i] == input_num)
            {
                printf("The number you entered is in the array");
                break;
            }

            else
            {
                printf("The number you entered is not in the array");
                break;
            }
        }
    }

in this for loop,if your input not match the arraylength[0],

then your method immediately print 'the number you entered is not in the array'

and then break the for loop,so you never get the chance to compare the rest of your array to input_num

Upvotes: 0

roottraveller
roottraveller

Reputation: 8232

your logic to check-in array is wrong. try this

void arrayresult(int arraylength[], int input_num) {  
    bool isFound = false;

   // iterate over till you found the  number or reaches end of array
    for (int i = 0; i < 10; i++){
        if (arraylength[i] == input_num){
            isFound = true;
            break;
        }
     }

     if(isFound == true) {
         printf("The number you entered is in the array");
      }else{
            printf("The number you entered is not in the array");
        }

}

Upvotes: 1

Venkatesh Nandigama
Venkatesh Nandigama

Reputation: 518

You can only conclude the number is not the list, after checking with all elements remove else from for loop. 

    else
        {
            printf("The number you entered is not in the array");
            break;
        }

write printf("not found") after the loop, and write return in if

Upvotes: 0

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